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(A) The given differential equation is $\sec x \, dy = -\{2(1-x) 	an x + x(2-x)\} \, dx$. Dividing by $\sec x$,we get $\frac{dy}{dx} = -\{2(1-x) \sin

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$2$

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(A) The given differential equation is $\sec x \, dy = -\{2(1-x) 	an x + x(2-x)\} \, dx$. Dividing by $\sec x$,we get $\frac{dy}{dx} = -\{2(1-x) \sin x + x(2-x) \cos x\}$. This simplifies to $\frac{dy}{dx} = 2(x-1) \sin x + (x^2-2x) \cos x$. Integrating both sides with respect to $x$: $y(x) = \int 2(x-1) \sin x \, dx + \int (x^2-2x) \cos x \, dx$. Using integration by parts on the second term: $\int (x^2-2x) \cos x \, dx = (x^2-2x) \sin x - \int (2x-2) \sin x \, dx$. Substituting this back: $y(x) = \int 2(x-1) \sin x \, dx + (x^2-2x) \sin x - \int 2(x-1) \sin x \, dx + C$. Thus,$y(x) = (x^2-2x) \sin x + C$. Given $y(0)=2$,we have $2 = (0^2-2(0)) \sin(0) + C$,which implies $C=2$. So,$y(x) = (x^2-2x) \sin x + 2$. For $x=2$,$y(2) = (2^2-2(2)) \sin(2) + 2 = (4-4) \sin(2) + 2 = 2$.

Let $y=y(x)$ be the solution of the differential equation $\sec x \, dy + \{2(1-x) \tan x + x(2-x)\} \, dx = 0$ such that $y(0)=2$. Then $y(2)$ is equal to :

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